【POJ 2728 Desert King】
Time Limit: 3000MS
Memory Limit: 65536K
Total Submissions: 27109
Accepted: 7527
Description
David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be watered. As the dominate ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way.
Input
There are several test cases. Each test case starts with a line containing a number N (2 <= N <= 1000), which is the number of villages. Each of the following N lines contains three integers, x, y and z (0 <= x, y < 10000, 0 <= z < 10000000). (x, y) is the position of the village and z is the altitude. The first village is the capital. A test case with N = 0 ends the input, and should not be processed.
Output
For each test case, output one line containing a decimal number, which is the minimum ratio of overall cost of the channels to the total length. This number should be rounded three digits after the decimal point.
Sample Input
4 0 0 0 0 1 1 1 1 2 1 0 3 0
Sample Output
1.000
Source
Beijing 2005
【翻譯】題目大意：有n個村莊要連在一起，村與村之間的長度為他們之間的水平距離，連在一起的花費是兩村的高度差。現求所花費和與長度和之比最小。
分析：
①Sigma的比值可以想到使用01分數規劃。
②01分數規劃有兩種實現方式：
(1)dichotomy[有點慢] (2)Dinkelbach[超級快]
③所以整個程序就是01分數規劃然後使用Prim求出最小生成樹。
④Prim很快，代碼又好寫。
#include<math.h>
#include<stdio.h>
#define go(i,a,b) for(int i=a;i<=b;i++)
#define G(i,j) (g[i][j].cost-r1*g[i][j].len)
const int N=1010;int vis[N];
double Sqr(double a){return a*a;};
double Abs(double a){return a<0?-a:a;}
struct E{double cost,len;}g[N][N],e[N];
int n,x[N],y[N],h[N],T=-1e9;double d[N],r1,r2;
double Dis(int i,int j){return sqrt(Sqr(x[i]-x[j])+Sqr(y[i]-y[j]));}
void Prim()
{
double C=0,D=0;d[1]=1e15;int v;++T;
go(u,2,n)d[u]=G(1,u),e[u]=g[1][u];go(i,2,n){v=1;
go(u,2,n)if(vis[u]!=T&&d[u]<d[v])v=u;vis[v]=T;C+=e[v].cost;
go(u,2,n)if(vis[u]!=T&&G(v,u)<d[u])d[u]=G(v,u),e[u]=g[v][u];D+=e[v].len;}
r2=C/D;
}
int main()
{
while(scanf("%d",&n),n)
{
go(i,1,n)scanf("%d%d%d",x+i,y+i,h+i);
go(i,1,n)go(j,1,n)g[i][j]=(E){Abs(h[i]-h[j]),Dis(i,j)};
r1=r2=0;while(Prim(),Abs(r1-r2)>1e-5)r1=r2;printf("%.3f\n",r1);
}
return 0;
}//Paul_Guderian
And every young mammal has multitudinous opportunities.————Judy·Hopps
【POJ 2728 Desert King】